<h1>Mark 解題紀錄</h1>
<h3>CPE一星必考題</h3>
1.<h3>1.Vito's Family </h3>
<p>題目大意：求出黑幫老大家到所有的親戚的家的距離的和為最小。</p>
```c++=
#include <bits/stdc++.h>
using namespace std;
int main()
{
int t=0;
int a[500]={0};
cin>>t;
while(t--){
int s=0;
cin>>s;
for(int i=0;i<s;i++){
cin>>a[i];
}
sort(a,a+s);
int mid=a[s/2];
int sum=0;
for(int i=0;i<s;i++){
sum+=abs(mid-a[i]);
}
cout<<sum<<endl;
}
}
```
---
<h3>2.Hashmat the brave warrior </h3>
<p>題目大意：計算己方與敵方士兵的數目差距。</p>
```c++=
#include <bits/stdc++.h>
using namespace std;
int main()
{
long long a,b;
while(cin>>a>>b){
if(a>b){
swap(a,b);
}
cout<<b-a<<endl;
}
}
```
---
<h3>3.Primary Arithmetic </h3>
<p>題目大意：由右至左一位一位相加。如果相加的結果大於等於10就有進位（carry）的情況出現。</p>
```c++=
#include <iostream>
using namespace std;
int main()
{
long long a,b;
while(cin>>a>>b){
int count=0;
int e=0;
if(a==0 && b==0){
break;
}else{
while(a!=0 || b!=0){
int c=a%10;
int d=b%10;
if(c+d+e>=10){
count++;
e=1;
}else{
e=0;
}
a=a/10;
b=b/10;
}
if(count==1){
cout<<"1 carry operation."<<endl;
}else if(count==0){
cout<<"No carry operation."<<endl;
}else{
cout<<count<<" carry operations."<<endl;
}
}
}
}
```
---
<h3>4.The 3n + 1 problem </h3>
<p>題目大意：輸入n,如果 n = 1 結束,如果 n 是奇數那麼n=3*n+1,此數列的長度稱為n的cycle-length。上面提到的例子, 22 的 cycle length為 16,問題來了：對任2個整數i，j我們想要知道介於i，j（包含i，j）之間的數所產生的數列中最大的 cycle length 是多少。</p>
```c++=
#include <bits/stdc++.h>
using namespace std;
int len(int i){
int lengs=0;
for(lengs=1;i!=1;lengs++){
if(i%2==0){
i/=2;
}else{
i=3*i+1;
}
}
return lengs;
}
int main()
{
long long a,b;
while(cin>>a>>b){
int max=0;
int start=0,end=0;
if(a>b){
start=b;
end=a;
}else{
start=a;
end=b;
}
for(int i=start;i<=end;i++){
if(len(i)>max){
max=len(i);
}
}
cout<<a<<" "<<b<<" "<<max<<endl;
}
}
```
---
<h3>5. You can say 11 </h3>
<p>題目大意：給你一個正整數 N，判定它是否是 11 的倍數。</p>
```c++=
#include <bits/stdc++.h>
using namespace std;
int main()
{
string s;
while(cin>>s){
int odd=0,two=0;
int sum=0;
if(s=="0"){
break;
}else{
for(int i=0;i<s.length();i+=2){
odd+=s[i]-'0';
}
for(int i=1;i<s.length();i+=2){
two+=s[i]-'0';
}
sum=abs(odd-two);
}
if(sum%11==0){
cout<<s<<" is a multiple of 11."<<endl;
}else{
cout<<s<<" is not a multiple of 11."<<endl;
}
}
}
```
---
<h3>6. Bangla Numbers </h3>
<p>題目大意：???</p>
```c++=
#include <bits/stdc++.h>
using namespace std;
void number(long long int a){
if(a>=10000000){
number(a/10000000);
cout<<" kuti";
a%=10000000;
}
if(a>=100000){
number(a/100000);
cout<<" lakh";
a%=100000;
}
if(a>=1000){
number(a/1000);
cout<<" hajar";
a%=1000;
}
if(a>=100){
number(a/100);
cout<<" shata";
a%=100;
}
if(a){
cout<<" "<<a;
}
}
int main()
{
long long int a;
int count=1;
while(cin>>a){
printf("%4d.",count);
if(a==0)cout<<" 0";
number(a);
cout<<endl;
count++;
}
}
```
---
<h3>7. List of Conquests </h3>
<p>題目大意：????。</p>
```c++=
#include <bits/stdc++.h>
using namespace std;
int main()
{
int a;
cin>>a;
string s;
string country[a];
for(int i=0;i<a;i++){
cin>>country[i];
getline(cin,s);
}
sort(country,country+a);
for (int i=0;i<a;){
int j=0;
cout << country[i] << " ";
int count = 1;
for (j=i+1;j<a;j++){
if (country[i] != country[j]){
break;
}
count++;
}
i = j;
cout << count << endl;
}
}
```
---
<h3>8. What's Cryptanalysis? </h3>
<p>題目大意：輸入一段文字，分析其 ASCII 與次數,單測資,輸入的第一個數字 n 代表接下來的輸入行數,每行輸入可能含有多個字元 (包含空白),輸入的字母大小寫視為同一字元,按照規則輸出 字母 (大寫) & 次數,按照次數由大到小,同樣次數按照 ASCII 由小到大</p>
```c++=
#include <bits/stdc++.h>
using namespace std;
int main()
{
int a;
cin>>a;
string s;
string country[a];
for(int i=0;i<a;i++){
cin>>country[i];
getline(cin,s);
}
sort(country,country+a);
for (int i=0;i<a;){
int j=0;
cout << country[i] << " ";
int count = 1;
for (j=i+1;j<a;j++){
if (country[i] != country[j]){
break;
}
count++;
}
i = j;
cout << count << endl;
}
}
```
---
<h3>9.Decode the Mad man </h3>
<p>題目大意：解碼,請看你鍵盤會發現規律是往減兩格就是答案</p>
```c++=
#include <bits/stdc++.h>
using namespace std;
int main()
{
string a="`1234567890-=qwertyuiop[]\\asdfghjkl;'zxcvbnm,./";
string s;
while(getline(cin,s)){
for(int i=0;i<s.length();i++){
if(s[i]<=90 && s[i]>=65){
s[i]+=32;
}
}
for(int i=0;i<s.length();i++){
if(s[i]==' '){
cout<<' ';
}
for(int j=0;j<a.length();j++){
if(s[i]==a[j]){
cout<<a[j-2];
}
}
}
cout<<endl;
}
}
```
---
<h3>10. Summing Digits</h3>
<p>題目大意：?????</p>
```c++=
#include <bits/stdc++.h>
using namespace std;
int count,b,n;
int number(int n){
b=0;
if(n>9){
while(n>0){
b+=n%10;
n/=10;
}
return number(b);
}else{
return n;
}
}
int main()
{
long long a;
while(cin>>a){
if(a==0){
break;
}else{
cout<<number(a)<<endl;
}
}
}
```
---
<h3>11. Common Permutation </h3>
<p>題目大意：給定兩個由小寫字母組成的字串a和b。印出最長的小寫字串x，使得x經過重新排列後為a的子序列，且x經過重新排列後為b的子序列。</p>
```c++=
#include <bits/stdc++.h>
using namespace std;
int main()
{
string s,t;
while(getline(cin,s)&&getline(cin,t)){
vector <char> a;
for(int i=0;i<s.length();i++){
for(int j=0;j<t.length();j++){
if(s[i]==t[j]){
a.push_back(s[i]);
t[j]='1';
break;
}
}
}
sort(a.begin(),a.end());
for(auto& i:a){
cout<<i;
}
cout<<endl;
}
}
```
---
<h3>12. Rotating Sentences </h3>
<p>題目大意：將數列文字往順時針方向旋轉90度。也就是說將原本由左到右，由上到下的句子輸出成由上到下，由右到左。</p>
```c++=
#include <bits/stdc++.h>
using namespace std;
int main(){
string s[101];
int coun=0,max=0;
while (getline(cin, s[coun])) {
if (s[coun].length()>max){
max=s[coun].length();
}
coun++;
}
for(int i=0;i<max;i++){
for(int j=coun-1;j>=0;j--){
if(s[j].length()>i){
cout<<s[j][i];
}else{
cout<<" ";
}
}
cout<<endl;
}
}
```
---
<h3>13. TEX Quotes </h3>
<p>題目大意：每一組雙引號的第一個 " 必須用``來代替,每一組雙引號的第二個 " 必須用''來代替。</p>
```c++=
#include <bits/stdc++.h>
using namespace std;
int main()
{
string s;
int count=1;
while(getline(cin,s)){
for(int i=0;i<s.length();i++){
if(s[i]=='"'){
if(count%2!=0){
cout<<"``";
count=2;
}else{
cout<<"''";
count=1;
}
}else{
cout<<s[i];
}
}
cout<<endl;
}
}
```
---
<h3>14. Doom's Day Algorithm</h3>
<p>題目大意：給你月份跟日期,求出那天是星期幾。</p>
```c++=
#include <bits/stdc++.h>
using namespace std;
int main()
{
int t;
cin>>t;
int Day[]={0,31,28,31,30,31,30,31,31,30,31,30,31};
while(t--){
int a,b,Today=0;
cin>>a>>b;
int days=0;
for(int i=1;i<a;i++){
days+=Day[i];
}
Today=(days+b)%7;
if(Today==1){
cout<<"Saturday"<<endl;
}else if(Today==2){
cout<<"Sunday"<<endl;
}else if(Today==3){
cout<<"Monday"<<endl;
}else if(Today==4){
cout<<"Tuesday"<<endl;
}else if(Today==5){
cout<<"Wednesday"<<endl;
}else if(Today==6){
cout<<"Thursday"<<endl;
}else if(Today==0){
cout<<"Friday"<<endl;
}
}
}
```
---
<h3>15. Jolly Jumpers</h3>
<p>題目大意：有n個整數的序列我們稱為jolly jumper，如果相鄰的2個數其差的絕對值恰好為1到n-1。</p>
```c++=
#include <bits/stdc++.h>
using namespace std;
int main(){
int a,b;
while(cin>>a){
int count=0;
vector <int> box;
vector <int> box2;
for(int i=0;i<a;i++){
cin>>b;
box.push_back(b);
}
for(int i=0;i<a-1;i++){
box2.push_back(abs(box[i]-box[i+1]));
}
sort(box2.begin(),box2.end());
int Count=0;
for(int i=0;i<a-1;i++){
if(box2[i]!=i+1){
Count=1;
break;
}
}
if(Count){
cout<<"Not jolly"<<endl;
}else{
cout<<"Jolly"<<endl;
}
}
}
```
---
<h3>16. What is the Probability ?</h3>
<p>題目大意：那玩家play的順序是從第一個到最後一個，若沒有人獲勝，就到第二輪，若第二輪還沒有人獲勝，就到第三輪，直到有人獲勝為止。算法大致上就是先求出第一輪x玩家獲勝的機率，再加上第二輪x玩家獲勝的機率...一直加到無窮大。</p>
```c++=
#include <bits/stdc++.h>
using namespace std;
int main()
{
int t,a,b;
double p,q,s;
cin>>t;
while(t--){
cin>>a>>p>>b;
q=1-p;
if(p==0)
cout<<"0.0000"<<endl;
else{
s=pow(q,b-1)*p/(1-pow(q,a));
printf("%.4f\n",s);
}
}
}
```
---
<h3>17. The Hotel with Infinite Rooms</h3>
<p>題目大意：如果一組四個成員的旅行團在8月1日早晨到達，則它將在8月4日晚上離開酒店下一組五個成員將在8月5日早晨入住並且停留五天，依此類推。現在給定第一組旅行團人數，您必須回答在指定日期入住的旅行團人數。</p>
```c++=
#include <bits/stdc++.h>
using namespace std;
int main()
{
long long s,d;
while(cin>>s>>d){
while(d>0){
d-=s;
if(d<=0){
break;
}else{
s++;
}
}
cout<<s<<endl;
}
}
```
---
<h3>18. 498-bis</h3>
<p>題目大意：給你一組方程式和X求微分後的值。</p>
```c++=
#include <bits/stdc++.h>
using namespace std;
int main()
{
int t;
string s;
while(cin>>t){
int sum=0;
vector <int> v;
getline(cin,s);
getline(cin,s);
stringstream ss(s);
int pw=1;
while(ss>>s){
v.push_back(stoi(s));
}
for(int i=v.size()-2;i>=0;i--){
sum+=v[i]*(v.size()-1-i)*pw;
pw*=t;
}
cout<<sum<<endl;
}
}
```
---
<h3>19. Odd Sum</h3>
<p>題目大意：給你一個範圍 a 到 b ，請你找出 a 與 b 之間所有奇數的和。</p>
```c++=
#include <bits/stdc++.h>
using namespace std;
int main()
{
int a;
int b=0,c=0;
int count=0;
cin>>a;
while(a--){
count++;
cin>>b>>c;
int sum=0;
for(int i=b;i<=c;i++){
if(i%2!=0){
sum+=i;
}
}
cout<<"Case "<<count<<": "<<sum<<endl;
}
}
```
---
<h3>20. Beat the Spread!</h3>
<p>題目大意：？？？？？</p>
```c++=
#include <bits/stdc++.h>
using namespace std;
int main() {
int t,a,b;
cin>>t;
while(t--){
cin>>a>>b;
if(b>a || (a+b)%2){
cout<<"impossible"<<endl;
}else{
cout<<(a+b)/2<<" "<<(a-b)/2<<endl;
}
}
}
```
---
<h3>21. Symmetric Matrix</h3>
<p>題目大意：？？？？？</p>
```c++=
#include <bits/stdc++.h>
using namespace std;
int main() {
}
```
---
<h3>22. Square Numbers</h3>
<p>題目大意：求出 a 與 b 之間 (含) 有幾個完全平方數。</p>
```c++=
#include <bits/stdc++.h>
using namespace std;
int main()
{
int a,b,sq;
while(cin>>a>>b && a!=0 && b!=0){
int count=0;
for(int i=a;i<=b;i++){
sq=sqrt(i);
if(sq*sq==i){
count++;
}
}
cout<<count<<endl;
}
}
```
---
<h3>23. B2-Sequence</h3>
<p>題目大意：「B2數列」係指一正整數數列 1<= b1 < b2 < b3 ...，其中所有的 bi + bj （i <= j）皆不相等。</p>
```c++=
#include <bits/stdc++.h>
using namespace std;
int main()
{
int t;
int count=0;
int a[t]={0};
while(cin>>t){
count++;
bool Hello=true;
for(int i=0;i<t;i++){
cin>>a[i];
if(a[i]<1){
Hello=false;
}
}
for(int i=0;i<t-1;i++){
if(a[i]>a[i+1]){
Hello=false;
}
}
int sum=0;
set <int> box;
box.clear();
if(Hello){
for(int i=0;i<t;i++){
for(int j=i;j<t;j++){
box.insert(a[i]+a[j]);
}
}
for(int i=0;i<=t;i++){
sum+=i;
}
if(box.size()==sum){
cout<<"Case #"<<count<<": It is a B2-Sequence."<<endl;
cout<<endl;
}else{
cout<<"Case #"<<count<<": It is not a B2-Sequence."<<endl;
cout<<endl;
}
}else{
cout<<"Case #"<<count<<": It is not a B2-Sequence."<<endl;
cout<<endl;
}
}
}
```
---
<h3>24. Back to High School Physics</h3>
<p>題目大意:求 2t 秒後所經過的位移是多少</p>
```c++=
#include <bits/stdc++.h>
using namespace std;
int main()
{
int a,b;
while(cin>>a>>b){
cout<<2*a*b<<endl;
}
}
```
---
<h3>25. An Easy Problem!</h3>
<p>題目大意:???</p>
```c++=
#include <bits/stdc++.h>
using namespace std;
int main()
{
string s;
while(getline(cin,s)){
int sum=0,max=1;
int test;
for(int i=0;i<s.length();i++){
if(s[i]>=48 && s[i]<=122){
if(s[i]>='0' && s[i]<='9'){
test=s[i]-48;
}else if(s[i]>='A' && s[i]<='Z'){
test=s[i]-55;
}else if(s[i]>='a' && s[i]<='z'){
test=s[i]-61;
}
sum+=test;
if(test>max){
max=test;
}
}else{
continue;
}
}
int i=0;
for(i=max;i<62;i++){
if(sum%i==0){
cout<<i+1<<endl;
break;
}
}
if(i==62){
cout<<"such number is impossible!"<<endl;
}
}
}
```
---
<h3>26. Fibonaccimal Base</h3>
<p>題目大意:給你一個十進位將他轉為費氏數列進制輸出。</p>
```c++=
#include <bits/stdc++.h>
using namespace std;
int main()
{
int FB[39]={0,1,1};
for(int i=3;i<=40;i++){
FB[i]=FB[i-1]+FB[i-2];
}
int t,a;
cin>>t;
while(t--){
vector <int> v;
cin>>a;
int b=a;
for(int i=39;i>1;i--){
if(a>=FB[i]){
a-=FB[i];
v.push_back(1);
}else{
v.push_back(0);
}
}
cout<<b<<" = ";
for(int i=0;i<=39;i++){
if(v[i]==1){
for(int j=i;j<=37;j++){
cout<<v[j];
}
break;
}
}
cout<<" (fib)"<<endl;
}
}
```
---
<h3>27. Funny Encryption Method</h3>
<p>題目大意:將十進制轉為二進制,算有幾個1,再將數字從十六進制轉為二進制,算有幾個1。</p>
```c++=
#include <bits/stdc++.h>
using namespace std;
int main(){
int t,a, b, c, d;
cin >> t;
while(t--){
int n=0,n1=0;
cin>>a;
int e=a;
while(a > 0){
b=a%2;
if(b==1){
n++;
}
a/=2;
}
while(e>0){
c=e%10;
while(c>0){
d=c%2;
if(d==1){
n1++;
}
c/=2;
}
e/=10;
}
cout<<n<< " " <<n1<<endl;
}
}
```
---
<h3>28. Parity</h3>
<p>題目大意:將十進制轉為二進制,算有幾個1。</p>
```c++=
#include <bits/stdc++.h>
using namespace std;
int main()
{
int a;
while(cin>>a && a){
int count=0;
vector <int> v;
while(a>0){
if(a%2==0){
v.push_back(0);
}else{
count++;
v.push_back(1);
}
a/=2;
}
cout<<"The parity of ";
reverse(v.begin(),v.end());
for(auto&i:v){
cout<<i;
}
cout<<" is "<<count<<" (mod 2)."<<endl;
}
}
```
---
<h3>29. Cheapest Base</h3>
<p>題目大意:將十進制轉為二進制,算有幾個1。</p>
```c++=
#include <bits/stdc++.h>
using namespace std;
int main()
{
int t,n,b;
int coun=0;
cin>>t;
while(t--){
coun++;
int cost[36]={0};
for(int i=0;i<36;i++){
cin>>cost[i];
}
cin>>n;
cout<<"Case "<<coun<<":"<<endl;
while(n--){
cin>>b;
int min=0,c,d;
int cheast[37]={0};
for(int i=2;i<=36;i++){
c=b;
int sum=0;
while(c>0){
d=c%i;
sum+=cost[d];
c/=i;
}
if(sum<=min || min==0)
{
cheast[i]=sum;
min=sum;
}
}
cout<<"Cheapest base(s) for number "<<b<<":";
for(int i=2;i<=36;i++){
if(cheast[i]==min){
cout<<" "<<i;
}
}
cout<<endl;
if(n==0 &&t!=0){
cout<<endl;
}
}
}
}
```
---
<h3>30. Hartals</h3>
<p>題目大意:?????</p>
```c++=
#include <bits/stdc++.h>
using namespace std;
int main()
{
int t,p,n;
cin>>t;
while(t--){
int a;
vector <int> v;
cin>>p;
cin>>n;
for(int i=0;i<n;i++){
cin>>a;
v.push_back(a);
}
int h=0;
for(int i=0;i<=p;i++){
if(i%7==6 || i%7==0){
continue;
}else{
for(int j=0;j<n;j++){
if(i%v[j]==0){
h++;
break;
}
}
}
}
cout<<h<<endl;
}
}
```
---
<h3>31. All You Need Is Love</h3>
<p>題目大意:看給的兩個二進位是不是互質</p>
```c++=
#include <bits/stdc++.h>
using namespace std;
int main()
{
int t;
string s,s1;
cin>>t;
int count=1;
while(t--){
cin>>s>>s1;
int sum=0;
int sum2=0;
for(int i=0;i<s.length();i++){
if(s[i]=='1'){
sum+=pow(2,s.length()-i-1);
}
}
for(int i=0;i<s1.length();i++){
if(s1[i]=='1'){
sum2+=pow(2,s1.length()-i-1);
}
}
if(__gcd(sum,sum2)!=1){
cout<<"Pair #"<<count<<": All you need is love!"<<endl;
}else{
cout<<"Pair #"<<count<<": Love is not all you need!"<<endl;
}
count++;
}
}
```
---
<h3>32. Divide, But Not Quite Conquer!</h3>
<p>題目大意:
1. a[1] = n, a[i] = a[i − 1] ÷ m, for all 1 < i ≤ k
2. a[i] 被 m 整除(a[i] mod m = 0) for all 1 ≤ i < k
3. a[1] > a[2] > a[3] > ... > a[k]
</p>
```c++=
#include <bits/stdc++.h>
using namespace std;
int main()
{
long long s,d;
while(cin>>s>>d){
if(d<=1 || s<d){
cout<<"Boring!"<<endl;
continue;
}
vector <long long int> box;
while(s){
box.push_back(s);
s/=d;
}
bool hello=true;
for(int i=0;i<box.size();i++){
if(i==box.size()-1){
break;
}
if(box[i]%d!=0){
hello=false;
}
}
for(int i=0;i<box.size()-1;i++){
if(box[i]<box[i+1]){
hello=false;
}
}
if(box.back()!=1){
hello=false;
}
if(hello){
for(int i = 0;i < box.size();i++){
cout<<box[i];
if(i!=box.size()-1){
cout<<" ";
}
}
cout<<endl;
}else{
cout<<"Boring!"<<endl;
}
}
}
```
---
<h3>33. Simply Emirp</h3>
<p>題目大意:看是不是互質,如果是互質在看倒數是不是互質。</p>
```c++=
#include <bits/stdc++.h>
using namespace std;
int main()
{
ios::sync_with_stdio(0);
cin.tie(0);
cout.tie(0);
int a;
while(cin>>a){
int count=1;
int b=0,c=0,re=0;
int d=a;
for(int i=2;i<a;i++){
if(a%i==0){
count=0;
break;
}
}
if(count==1){
while(a!=0){
b=a%10;
re=re*10+b;
a/=10;
}
if(re!=d){
count=2;
for(int i=2;i<re;i++){
if(re%i==0){
count=1;
break;
}
}
}
}
if(count==2){
cout<<d<<" is emirp."<<endl;
}else if(count==1){
cout<<d<<" is prime."<<endl;
}else{
cout<<d<<" is not prime."<<endl;
}
}
}
```
---
<h3>34. 2 the 9s</h3>
<p>題目大意:給你一個正整數N，判斷他是不是9的倍數，而且如果他是9的倍數你還需要判斷它的 9-degree。</p>
```c++=
#include <bits/stdc++.h>
using namespace std;
int count;
int number(long long int n){
if(n % 9) return 0;
else if(n == 9){
return ++count;
}
int i = 0;
while(n){
i += n % 10;
n /= 10;
}
count++;
return number(i);
}
int main()
{
string s;
while(cin>>s && s !="0"){
int sum=0;
for(int i=0;i<s.length();i++){
sum+=s[i]-'0';
}
count = 0;
count = number(sum);
if(count!=0){
cout<<s<<" is a multiple of 9 and has 9-degree "<<count <<".\n";
}else{
cout<<s<<" is not a multiple of 9.\n";
}
}
}
```
---
<h3>35. GCD</h3>
<p>題目大意:找出因數。</p>
```c++=
#include <bits/stdc++.h>
using namespace std;
int main()
{
int t;
while(cin>>t &&t){
int G=0;
for(int i=1;i<t;i++){
for(int j=i+1;j<=t;j++){
G+=__gcd(i,j);
}
}
cout<<G<<endl;
}
return 0;
}
```
---
<h3>36. Largest Square</h3>
<p>題目大意:==。</p>
```c++=
我不會寫
```
---
<h3>37. Satellites</h3>
<p>題目大意:==。</p>
```c++=
#include <bits/stdc++.h>
using namespace std;
int main()
{
double s,a;
string b;
while(cin>>s>>a>>b){
double pi=3.14159265359/180;
if(b[0]=='m'){
a/=60;
}
while(a>=360){
a-=360;
}
while(a>180){
abs(a-=360);
}
printf("%.6f %.6f\n",abs((s+6440)*a*pi),abs((s+6440)*sin(a/2*pi)*2));
}
}
```
---
<h3>38. Can You Solve It?</h3>
<p>題目大意:花費的最小步數。</p>
```c++=
#include <bits/stdc++.h>
using namespace std;
int number(int x,int y){
return (x+y)*(x+y+1)/2+x;
}
int main()
{
int n,x1,y1,x2,y2;
int coun=0;
cin>>n;
while(n--){
coun++;
cin>>x1>>y1>>x2>>y2;
int count=0;
cout<<"Case "<<coun<<": "<<number(x2,y2)-number(x1,y1)<<endl;
}
}
```
---
<h3>39. Fourth Point !!</h3>
<p>題目大意:給定平行四邊形的兩個相鄰邊的端點的 (x, y)座標,找到第四個點的 (x, y)座標。</p>
```c++=
#include <bits/stdc++.h>
using namespace std;
int main()
{
double x[4],y[4],ax,ay;
while(cin>>x[0]>>y[0]){
ax=x[0];
ay=y[0];
for(int i=1;i<4;i++){
cin>>x[i]>>y[i];
ax+=x[i];
ay+=y[i];
}
for(int i=0;i<4;i++){
for(int j=i+1;j<4;j++){
if(x[i]==x[j] && y[i]==y[j]){
cout<<fixed<<setprecision(3)<<ax-3*x[i]<<' '<<ay-3*y[i]<<endl;
}
}
}
}
}
```
---
<h3>40. A mid-summer night's dream.</h3>
<p>題目大意:????。</p>
```c++=
#include <bits/stdc++.h>
using namespace std;
int main()
{
int n;
int a;
while(cin>>n){
vector <int> v;
for(int i=0;i<n;i++){
cin>>a;
v.push_back(a);
}
sort(v.begin(),v.end());
int mid=0,mid2=0;
if(n%2==0){
mid=v[n/2-1];
mid2=v[n/2];
}else{
mid=v[n/2];
mid2=mid;
}
int coun=0;
for(int i=0;i<v.size();i++){
if(v[i]==mid || v[i]==mid2){
coun++;
}
}
cout<<mid<<" "<<coun<<" "<<(mid2-mid)+1<<endl;
}
}
```
---
<h3>41. Tell me the frequencies!</h3>
<p>題目大意:給你一列文字，請你找出各字元出現的次數。</p>
```c++=
#include <bits/stdc++.h>
using namespace std;
int main()
{
string s;
int coun=0;
while(getline(cin,s)){
if(coun++){
cout<<endl;
}
int count[200]={0};
for(int i=0;i<s.length();i++){
count[s[i]]++;
}
for(int i=1;i<=s.length();i++){
for(int j=127;j>= 32;j--){
if(count[j]==i){
cout<<j<<" "<<i<<endl;
}
}
}
}
}
```
---
<h3>42. Train Swapping</h3>
<p>題目大意:計算最少需要交換幾次兩個相鄰車廂，才能將所有車廂依序排好。</p>
```c++=
#include <bits/stdc++.h>
using namespace std;
int main()
{
int t;
cin>>t;
while(t--){
int n=0,count=0;
cin>>n;
int a[50]={0};
for(int i=0;i<n;i++){
cin>>a[i];
}
for(int i=0;i<n-1;i++){
for(int j=0;j<n-1-i;j++){
if(a[j]>a[j+1]){
int temp=a[j];
a[j]=a[j+1];
a[j+1]=temp;
count++;
}
}
}
cout<<"Optimal train swapping takes "<<count<<" swaps."<<endl;
}
}
```
---
<h3>43. Hardwood Species</h3>
<p>題目大意:給你種子名稱,算出種子占幾趴</p>
```c++=
#include <bits/stdc++.h>
using namespace std;
int main()
{
int t;
string s,a;
cin>>t;
getline(cin, a);
getline(cin, a);
while(t--){
double sum=0;
map <string , int> m;
while(getline(cin,s) && !s.empty()){
m[s]++;
sum++;
}
for(auto&i:m){
cout<<i.first<<" "<<fixed<<setprecision(4)<<(i.second/sum)*100<<endl;
}
if(t>0){
cout<<endl;
}
}
}
```
---
<h3>44. Minesweeper</h3>
<p>題目大意:算那九宮格周遭出現幾次地雷。</p>
```c++=
#include <bits/stdc++.h>
using namespace std;
int main()
{
int a,b,qq=0;
int count=0;
while(cin>>a>>b && (a || b)){
if(qq++){
cout<<endl;
}
char c;
count++;
int d[102][102]={0};
for(int i=1;i<=a;i++){
for(int j=1;j<=b;j++){
cin>>c;
if(c=='*'){
d[i][j]-=10;
for(int x=i-1;x<=i+1;x++){
for(int y=j-1;y<=j+1;y++){
d[x][y]++;
}
}
}
}
}
cout<<"Field #"<<count<<":"<<endl;
for(int i=1;i<=a;i++){
for(int j=1;j<=b;j++){
if(d[i][j]<0){
cout<<'*';
}else{
cout<<d[i][j];
}
}
cout<<endl;
}
}
}
```
---
<h3>45. Die Game</h3>
<p>題目大意:告訴你方向,輸出點數。</p>
```c++=
#include <bits/stdc++.h>
using namespace std;
int main()
{
map <string ,int> mp;
int a,temp;
string s;
while(cin>>a && a){
mp["top"]=1;
mp["north"]=2;
mp["west"]=3;
mp["east"]=4;
mp["south"]=5;
mp["bottom"]=6;
while(a--){
cin>>s;
if(s=="north"){
temp=mp["north"];
mp["north"]=mp["top"];
mp["top"]=mp["south"];
mp["south"]=mp["bottom"];
mp["bottom"]=temp;
}else if(s=="east"){
temp=mp["east"];
mp["east"]=mp["top"];
mp["top"]=mp["west"];
mp["west"]=mp["bottom"];
mp["bottom"]=temp;
}else if(s=="south"){
temp=mp["south"];
mp["south"]=mp["top"];
mp["top"]=mp["north"];
mp["north"]=mp["bottom"];
mp["bottom"]=temp;
}else if(s=="west"){
temp=mp["west"];
mp["west"]=mp["top"];
mp["top"]=mp["east"];
mp["east"]=mp["bottom"];
mp["bottom"]=temp;
}
}
cout<<mp["top"]<<endl;
}
}
```
---
<h3>46. Eb Alto Saxophone Player</h3>
<p>題目大意:計算每一根手指頭按了多少次按鍵,如果某一按鍵在下一音符時不會用到,那麼就會放開,否則就是維持按著的情況。</p>
```c++=
#include <bits/stdc++.h>
using namespace std;
int main()
{
map <char, vector<int>> mp;
mp['a']={0,1,1,0,0,0,0,0,0,0};
mp['b']={0,1,0,0,0,0,0,0,0,0};
mp['c']={0,1,1,1,0,0,1,1,1,1};
mp['d']={0,1,1,1,0,0,1,1,1,0};
mp['e']={0,1,1,1,0,0,1,1,0,0};
mp['f']={0,1,1,1,0,0,1,0,0,0};
mp['g']={0,1,1,1,0,0,0,0,0,0};
mp['A']={1,1,1,0,0,0,0,0,0,0};
mp['B']={1,1,0,0,0,0,0,0,0,0};
mp['C']={0,0,1,0,0,0,0,0,0,0},
mp['D']={1,1,1,1,0,0,1,1,1,0};
mp['E']={1,1,1,1,0,0,1,1,0,0};
mp['F']={1,1,1,1,0,0,1,0,0,0};
mp['G']={1,1,1,1,0,0,0,0,0,0};
int t;
string s;
cin>>t;
getline(cin,s);
while(t--){
getline(cin,s);
int ans[11]={0};
int last[11]={0};
for(int i=0;i<s.length();i++){
for(int j=0;j<10;j++){
if(mp[s[i]][j]){
if (last[j]){
continue;
}
else {
last[j] = 1;
ans[j]++;
}
}else{
last[j]=0;
}
}
}
for(int i=0;i<10;i++){
cout<<ans[i];
if(i!=9){
cout<<" ";
}
}
cout<<endl;
}
}
```
---
<h3>47. Mutant Flatworld Explorers</h3>
<p>題目大意:計算每一根手指頭按了多少次按鍵,如果某一按鍵在下一音符時不會用到,那麼就會放開,否則就是維持按著的情況。</p>
```c++=
#include <bits/stdc++.h>
using namespace std;
int main()
{
map <char, vector<int>> mp;
mp['a']={0,1,1,0,0,0,0,0,0,0};
mp['b']={0,1,0,0,0,0,0,0,0,0};
mp['c']={0,1,1,1,0,0,1,1,1,1};
mp['d']={0,1,1,1,0,0,1,1,1,0};
mp['e']={0,1,1,1,0,0,1,1,0,0};
mp['f']={0,1,1,1,0,0,1,0,0,0};
mp['g']={0,1,1,1,0,0,0,0,0,0};
mp['A']={1,1,1,0,0,0,0,0,0,0};
mp['B']={1,1,0,0,0,0,0,0,0,0};
mp['C']={0,0,1,0,0,0,0,0,0,0},
mp['D']={1,1,1,1,0,0,1,1,1,0};
mp['E']={1,1,1,1,0,0,1,1,0,0};
mp['F']={1,1,1,1,0,0,1,0,0,0};
mp['G']={1,1,1,1,0,0,0,0,0,0};
int t;
string s;
cin>>t;
getline(cin,s);
while(t--){
getline(cin,s);
int ans[11]={0};
int last[11]={0};
for(int i=0;i<s.length();i++){
for(int j=0;j<10;j++){
if(mp[s[i]][j]){
if (last[j]){
continue;
}
else {
last[j] = 1;
ans[j]++;
}
}else{
last[j]=0;
}
}
}
for(int i=0;i<10;i++){
cout<<ans[i];
if(i!=9){
cout<<" ";
}
}
cout<<endl;
}
}
```
---
<h3>48. Cola</h3>
<p>題目大意:3瓶空可樂罐換一瓶可樂。</p>
```c++=
#include <bits/stdc++.h>
using namespace std;
int main()
{
int a;
while(cin>>a){
cout<<(int)(1.5*a)<<endl;
}
}
```
---
<h3>49. Sort! Sort!! and Sort!!!</h3>
<p>題目大意:先利用每個數字除以M的餘數由小到大排，若排序中比較的兩數為一奇一偶且兩數除以M 的餘數相等，則奇數要排在偶數前面。若兩奇數除以M餘數大小相等，則原本數值較大的奇數排在前面。同樣的，若兩偶數除以M餘數大小相等，則較小的偶數排在前面。至於負數的餘數計算和 C 語言裡的定義相同，即負數的餘數絕對不會大於零。</p>
```c++=
#include <bits/stdc++.h>
using namespace std;
int n,m;
bool exe(int x,int y){
if((x%m) != (y%m)){
return (x%m)<(y%m);
}else{
if((x%2!=0) && (y%2!=0)){
return x>y;
}else if((x%2==0) && (y%2==0)){
return x<y;
}else{
return(x%2);
}
}
}
int main()
{
while(cin>>n>>m){
cout<<n<<" "<<m<<endl;
vector <int> s(n);
for(auto& i:s){
cin>>i;
}
sort(s.begin(), s.end(), exe);
for(auto& i:s){
cout<<i<<endl;
}
}
}
```
---
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